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79 lines
2.3 KiB
Plaintext
79 lines
2.3 KiB
Plaintext
Homework Day 8, Sept 21
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1) NGC 7027 has a distance of about 920 pc and diameter of about 14
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seconds of arc
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a. How does its diameter compare with its thickness (from previous
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homework)?
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Diameter D = θ(radians) × Distance d
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= 6.7787×10⁻⁵ radians × 920 pc
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= 0.0624 pc.
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This is very comparable to the result from HW7 (0.617). Well within 10%.
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b. Its shell is expanding at about 20km/s. How old is it? (how long
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ago was the shell ejected?)
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age t = (0.0624 pc) / (20 km s⁻¹ × 3.24×10⁻¹⁷ pc / 10⁻³ km)
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× (3.17×10⁻⁸ years / 1.00 s)
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= 4.50×10⁷ years.
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2) The Orion nebula is about 400 pc distant and has an apparent diameter of about 60 minutes of arc.
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a. What is its physical diameter?
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D = 6.98 pc.
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b. The gas density is about 10 4 cm -3 and its temperature is about 10 4 K. The turnover frequency is obvious on Wilson’s SED plot. How thick is the H + layer along our line of sight?
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turnover ν = 500 MHz.
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At turnover, τ = 1.
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Gaunt factor g = 10.6 + 1.9 log(T) - 1.26 log(Z ν)
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= 7.24.
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κ′ₛ = 0.178 Z² g ν⁻² T^(-3/2) nₑ nₕ
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= 5.15×10⁻¹⁶ cm⁻¹.
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Thickness l = τ / κ′ₛ.
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3) The Perseus cluster of galaxies is about 73.6 Mpc distant and has an angular diameter of 860 arc minutes. The total mass in stars is about 2×10 13 M sun and the virial mass is about 10 15 M sun . It is one of the strongest x-ray sources in the sky, with L x = 4x10 45 erg/s. X-rays come from 5 million Kelvin brems.
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a. Use total luminosity to find the emission measure.
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L = 4π j V.
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4π j = (From AGN3) 1.42×10⁻²⁷ Z² √T g[ff] n².
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I will adopt an average g[ff] = 1.3.
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Emission measure n² V = L / (4π 1.42×10⁻²⁷ Z² √T g[ff])
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= 7.71×10⁶⁷ cm⁻³.
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b. Use diameter to find V
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Spherical V = 4/3 π r³
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= 4/3 π (2.84×10²⁵ cm)³
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= 9.61×10⁷⁶ cm³.
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c. Solve for the density
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n² = 7.71×10⁶⁷ cm⁻³ / 9.61×10⁷⁶ cm³
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= 8.03×10⁻¹⁰.
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n = √n² = 2.83×10⁻⁵ cm⁻³.
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d. What is the total mass in hot gas?
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M = μ n V
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μₕ = 1.67×10⁻²⁷ kg × 2.83×10⁻⁵ cm⁻³ × 9.61×10⁷⁶ cm³
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= 2.29×10¹⁵ Mₛₒₗ. |